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With $N=m+n$ and no ties, sampling $m$ ranks without replacement gives the displayed mean and variance. For observed rank sum below its mean, a continuity-corrected lower-tail statistic is $Z=[W_X-m(N+1)/2+1/2]/\sqrt{mn(N+1)/12}$. Ties of sizes $t_g$ replace the variance by $mn[(N+1)-{\sum }_g(t_g^3-t_g)/(N(N-1))]/12$. These formulas concern the raw rank sum; the Mann–Whitney U test statistic subtracts $m(m+1)/2$ from both the observation and its mean.

The first output uses the equal-variance, pooled Student's t-test, as shown by its seven statistical degrees of freedom. Its exact model assumes two independent samples from normal distributions, with means ${\mu }_X,{\mu }_Y$ and a common unknown variance ${\sigma }^2$. The hypotheses are $H_0:{\mu }_X={\mu }_Y$ against $H_1:{\mu }_X<{\mu }_Y$. The sample means are $\bar{x}=4.35$ and $\bar{y}=9.88$, and the unbiased within-sample sample variances areThe pooled sample variance and its resulting standard error areHenceTo justify its reference distribution, under the null $(\bar{X}-\bar{Y})/(\sigma \sqrt{1/4+1/5})$ is standard normal, while $(3s_X^2+4s_Y^2)/{\sigma }^2$ is ${\chi }_7^2$, independently of the two sample means. Dividing the first variable by the square root of the second divided by seven gives the exact pooled two-sample t statistic, with Student t-distribution $t_7$. Its one-sided p-value is the lower tailThe associated one-sided 95% confidence interval for ${\mu }_X-{\mu }_Y$ isThe printed missing lower bound means that this is a one-sided upper confidence limit, not that a finite lower endpoint should be estimated. With such small samples the normality and common-variance assumptions matter; a Welch analysis would use a different variance estimate and generally different degrees of freedom.

The second output is the Wilcoxon rank sum test for two independent continuous samples. Under a common-distribution null hypothesis, labels are exchangeable conditional on the pooled values, so the four ranks assigned to the first sample form a uniformly chosen four-element subset of $\{1,\dots ,9\}$. The first sample receives ranks $2,1,3,5$, giving raw rank sumEquivalently the Mann–Whitney U test statistic is $U_X=W_X-4\cdot 5/2=1$, the number of cross-sample pairs with an $X$ observation larger than a $Y$ observation. This output reports the raw rank sum, rather than the shifted statistic used by some software.

A first sample tending to be smaller has a small rank sum. Under the null, all $\left(\genfrac{}{}{0ex}{}{9}{4}\right)=126$ rank allocations have equal probability. Only $\{1,2,3,4\}$ and $\{1,2,3,5\}$ have rank sum at most 11. By the exact rank-sum tail one above the minimum, the exact lower-tail p-value isNo large-sample normal approximation or continuity correction is involved; the absence of ties makes this simple enumeration valid. For a common-shape location-shift family, the test's alternative is a negative location shift of $X$ relative to $Y$. Without that model, the null is equality of distributions and the rank comparison is not simply a test of equality of means or medians.

Both tests reject at the 5% level in the direction of smaller $X$ observations; only the pooled t-test rejects at 1%. The t-test uses the numerical spacing and targets a mean difference under its normal model. The rank-sum test uses ordering and an exact exchangeability calculation, avoiding the normality and common-normal-variance assumptions. Their different p-values therefore reflect both different statistics and different modelling assumptions, not a numerical contradiction.

For the one-sided Wilcoxon rank sum test, suppose $X_1,\dots ,X_m$ and $Y_1,\dots ,Y_n$ are independent samples, iid within each sample, with continuous distributions. The null hypothesis is a common distribution. For a lower-tail alternative, the $X$ distribution is stochastically smaller. In the common-shape location model $X\stackrel{d}{=}Y+\delta$, this is $H_0:\delta =0$ against $H_1:\delta <0$; in that model it also compares medians. Without a location-family assumption, equality or inequality of medians alone is not the defining rank-sum hypothesis.

Rank all $N=m+n$ observations from smallest to largest and form $W_X={\sum }_{i=1}^m\mathrm{rank}(X_i)$. Reject for small $W_X$. Under $H_0$, the labels are exchangeable, so every subset of $m$ of the $N$ ranks is equally probable. The exact null distribution of a rank sum is thereforeEnumerate these subsets, or use the coefficient of $z^m{t}^w$ in ${\prod }_{k=1}^N(1+z{t}^k)$. An efficient recursion adds rank $k$ either to the selected sample or not: $a_{k,h,w}=a_{k-1,h,w}+a_{k-1,h-1,w-k}$, starting with $a_{0,0,0}=1$. The lower-tail P-value is ${\mathrm{Pr}}_0(W_X\le W_{\mathrm{obs}})$. A nonrandomized level-$\alpha$ test chooses a lower-tail rejection region whose null probability is at most $\alpha$.

The shifted Mann–Whitney U test statistic is $U_X=W_X-m(m+1)/2$. It counts pairs with $Y_j<X_i$, so the same lower-tail test uses small $U_X$. The distinction is important for these S-Plus outputs: the printed values are raw rank sums, not the shifted statistic used by some other software.

For the one-sided Wilcoxon signed-rank test, the pairs $(U_i,V_i)$ are independent between individuals but their two measurements may be dependent. Let $D_i=U_i-V_i$, and assume the differences are iid, continuous and symmetric about a common location $\delta$. Test $H_0:\delta =0$ against $H_1:\delta <0$. Rank the absolute differences and useUnder the null, conditional on their absolute values, the signs are independent fair signs. Every subset of ranks therefore has probability $2^{-n}$ of being the positive-sign subset, giving the exact null distribution of $W_{+}$. Reject for small $W_{+}$ and calculate the lower-tail probability. Its null mean and variance are $n(n+1)/4$ and $n(n+1)(2n+1)/24$. The symmetry assumption is essential: zero median by itself does not give this law. With zeros or ties, omit zero differences in the usual signed-rank version and use average ranks for ties, adjusting the distribution accordingly. The paired test is not appropriate for the independent RHR samples here.

For the experienced participants, let $X$ denote RHR1 running time and $Y$ denote RHR2 running time. The null hypothesis is identical running-time distributions; under the location interpretation it is zero RHR1-minus-RHR2 shift. The one-sided alternative is a negative shift, so RHR1 participants tend to run faster. With $m=n=13$, the printed statistic isAt $5\%$, reject the null in favour of faster RHR1 running times. The sample medians $11.52$ and $12.78$ minutes agree with that direction; their difference is descriptive, not the rank-sum test statistic.

For the new participants the same independent-sample Wilcoxon rank sum test compares eight RHR1 times against nine RHR2 times, again with a lower-tail alternative. HereThe exact null distribution of a rank sum gives this probability. It does not reject at $5\%$, although it is just below $10\%$. The data suggest the same direction but are less decisive; this is not evidence that the two new-participant distributions are equal.

The final command pools the experienced and new RHR1 samples into $m=21$ observations, and the corresponding RHR2 samples into $n=22$. It repeats the lower-tail rank-sum comparison on these two pooled groups. exact=F requests a normal approximation for a rank sum with a continuity correction. Without ties,The lower-tail statistic is $(W_X-462+1/2)/\sqrt{1694}$, with the usual tie-adjusted variance if ties occur. The supplied output iswhere $\Phi$ is the standard normal distribution function. Thus the pooled test rejects at $5\%$. The unreported pooled ranks must be formed afresh: the within-cohort raw rank sums cannot simply be added because pooling introduces cross-cohort comparisons.

Taken together, there is evidence that participants classified as RHR1 have shorter running times among the experienced men and in the pooled sample. The smaller new-participant samples give weaker evidence in the same direction. A significant result in one cohort and a nonsignificant result in another does not itself prove that their RHR effects differ. Pooling also assumes a meaningful pooled comparison; if programme duration changes the distributions, a cohort-adjusted analysis would be more informative. These observational comparisons do not establish that the fitness programme caused the differences, and none of the three tests directly compares experienced participants with new participants.